Definite Integral from 0 to 2 Pi of Reciprocal of One minus 2 a Cosine x plus a Squared
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Theorem
- $\ds \int_0^{2 \pi} \frac {\d x} {1 - 2 a \cos x + a^2} = \frac {2 \pi} {1 - a^2}$
where $a$ is a real number with $0 < a < 1$.
Proof
This article, or a section of it, needs explaining. In particular: The context of this needs to be explained a little more deeply: the integrand is defined as a real function, but the analysis is actually in the complex plane. The latter needs to be brought forward so as to make it clear to the reader. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
Let $C$ be the unit open disk centred at $0$.
The boundary of $C$, $\partial C$, can be parameterized by:
- $\map \gamma \theta = e^{i \theta}$
for $0 \le \theta \le 2 \pi$.
We have:
\(\ds \int_0^{2 \pi} \frac {\d x} {1 - 2 a \cos x + a^2}\) | \(=\) | \(\ds \int_0^{2 \pi} \frac {\d x} {1 - a \paren {e^{i x} + e^{-i x} } + a^2}\) | Euler's Cosine Identity | |||||||||||
\(\ds \) | \(=\) | \(\ds -\int_0^{2 \pi} \frac {e^{i x} } {a e^{2 i x} - \paren {a^2 + 1} e^{i x} + a} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 1 a \int_0^{2 \pi} \frac {e^{i x} } {e^{2 i x} - \paren {a + \frac 1 a} e^{i x} + \frac a a} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 1 a \int_0^{2 \pi} \frac {e^{i x} } {\paren {e^{i x} - \frac 1 a} \paren {e^{i x} - a} } \rd x\) | factorising | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 1 {i a} \int_{\partial C} \frac 1 {\paren {z - \frac 1 a} \paren {z - a} } \rd z\) | Definition of Complex Contour Integral, Derivative of Exponential Function |
- $z_1 = a$
and:
- $z_2 = \dfrac 1 a$
We have $0 < a < 1$, so $\dfrac 1 a > 1$.
So:
- $z_2$ lies outside the closed disk $\size z \le 1$
and:
- $z_1$ lies in $C$
Thus the only pole of concern is $z_1$.
Therefore:
\(\ds -\frac 1 {i a} \int_{\partial C} \frac 1 {\paren {z - \frac 1 a} \paren {z - a} } \rd z\) | \(=\) | \(\ds -\frac {2 \pi} a \Res {\frac 1 {\paren {z - \frac 1 a} \paren {z - a} } } a\) | Cauchy's Residue Theorem | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac {2 \pi} a \paren {\frac 1 {2 z - \paren {a + \frac 1 a} } }_{z \mathop = a}\) | Residue at Simple Pole, Derivative of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac {2 \pi} a \paren {\frac 1 {2 a - a - \frac 1 a} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \pi} {1 - a^2}\) |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Trigonometric Functions: $15.46$